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The number of views of a page on a Web site follows aPoisson distribution with a mean of 1.5 per minute.a. What is the probability of no views in a minute?b. What is the probability of two or fewer views in10 minutes?c. Does the answer to the previous part depend on whetherthe 10-minute period is an uninterrupted interval? Explain.d. Determine the length of a time interval such that the probabilityof no views in an interval of this length is 0.001.

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Answer:

a. P (X=0) = 0.223

b. P(X≤2) = 3.93 * 10^-5

C. No

D. t=4.6 miutes

Step-by-step explanation:

a. P (X=0) = [tex]e^{-1.5}[/tex] = 0.223

b. P(X≤2) =[tex]e^{1.5 * 10} (1 + 1.5*10 + \frac{1.5^{2} * 10^{2} }{2} )[/tex]

   = 3.93 * 10^-5

C. No, the answer to the previous part does not depend on whether the 10-minute period is an uninterrupted interval, it only depend on the uniformity of the density of views per minute ad independency of the disjoint time intervals.

D. P (X=0) =0.001

  [tex]e^{-1.5t} = 0.001\\\\-1.5t=ln(0.001)\\\\t =\frac{-6.9}{-1.5}[/tex]

t=4.6 miutes

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